Question

: Factorise x3 - 23x2 + 142x - 120.
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Answer 1

$\huge \fbox \red{ANSWER}$

x³- 23x² + 142x - 120

Some of the factors of -120 are: ±1, ±2, ±3, ±4, ±15, ±20, ±24, ±30, ±60

By trial, we put p(1)=0.

So, (x-1) is factor of p(x)

Now by solving this

= x³-23x+142x-120

= x³-x²-22x²+22x+120x-120

= x*(x-1)-22x(x-1)+120(x-1)

= (x-1)(x²-22x+120)...........(we took (x-1) common)

Now we get x²-22x+120

Now solving this by splitting the middle term we get

= x2-22x+120

= x2-12x-10x+120

= x(x-12)-10(x-12)

= (x-10)(x-12)

so the final answer by factoring the term

x³- 23x² + 142x - 120 we got= (x-1)(x-10)(x-12)

hope it helps you ☺️